Let $x_k=e^{\frac{2k\pi}{2n+1}}$ is a root of $x^{2n+1}=0$, where $k=0, 1, 2, \cdots, 2n$. $$(x-e^{\frac{k\cdot 2\pi}{2n+1}})(x-e^{\frac{(2n+1-k)\cdot2\pi}{2n+1}})= x^2 -2x\cos\frac{2k\pi}{2n+1}+1$$ Hence: $$(x^2 -2x\cos\frac{2\pi}{2n+1}+1)(x^2 -2x\cos\frac{4\pi}{2n+1}+1)\cdots(x^2 -2x\cos\frac{2n\pi}{2n+1}+1)=x^{2n}+x^{2n-1}+\cdots+x+1$$ Let $x=1 \implies (2-2x\cos\frac{2\pi}{2n+1})(2-2x\cos\frac{4\pi}{2n+1})\cdots(2-2x\cos\frac{2n\pi}{2n+1})=2n+1$ It follows:$$2^{2n}\sin^2\frac{\pi}{2n+1}\sin^2\frac{2\pi}{2n+1}\cdots\sin^2\frac{n\pi}{2n+1}=2n+1$$